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A person invested a total amount of Rs $15$ lakh. A part of it was invested in a fixed deposit earning $6\%$ annual interest, and the remaining amount was invested in two other deposits in the ratio $2:1$, earning annual interest at the rates of $4\%$ and $3\%$, respectively. If the total annual interest income is Rs $76000$ then the amount (in Rs lakh) invested in the fixed deposit was _______
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Given that,

A person invested a total amount of $ \text{Rs.} \; 15 \; \text{lakh} .$

A part of it he invested in a fixed deposit, and earns $6\%$ annual interest. 

Let $y$ lakh be the amount invested in a fixed deposit.

So, $ \text{Principle} = y, \;  \text{Rate} = 6\%$

The remaining invested amount $ = ( 15 – y )$

The remaining amount was invested in two deposits in the ratio $2:1,$ earning annual interest at the rates of $4\%$ and $3\%,$ respectively. 

  • First deposit $: \text{Principle} = \frac{2}{3} (15 – y), \; \text{Rate} = 4\% $
  • Second deposit $: \text{Principle} = \frac{1}{3} (15 – y), \; \text{Rate} = 3\%$

The total annual interest income $ = \text{Rs.} 76000 = \frac{76000}{100000} \; \text{(lakh)} = .76 \; \text{lakh}$ 

We know that$,\boxed{\text{Annual interest income} \propto \left(\text{Principle} \times \text{Rate}\right)}$

Now$,6\% \times y + \frac{2}{3}(15-y)4\% + \frac{1}{3}(15-y)3\% = .76$

$ \Rightarrow \frac{6y}{100} + \frac{8}{300}(15-y) + \frac{1}{100}(15-y) = .76 $

$ \Rightarrow \frac { 18y + 8(15-y) + 3(15-y)}{300} = .76 $

$ \Rightarrow 18y + 120 – 8y + 45 – 3y = \frac {76 \times 300}{100} $

$\Rightarrow 18y + 165 – 11y = 228 $

$ \Rightarrow 7y = 228 – 165 $

$ \Rightarrow 7y = 63 $

$ \Rightarrow \boxed{y = 9}$

$ \therefore$ The amount invested in the fixed deposit $ = 9 \; \text{lakh}.$

Correct Answer $: 9$

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